Mohr Circles and Conic Sections

What is the formula for the radius r of the ellipse at left?
The familiar Cartesian coordinate formula is shown below the ellipse.
We begin by noting that:

x = r cos A
y = r sin A

and substituting those formulas into the Cartesian equation.

When we do that substitution, we get:

r2cos2A/a2 + r2sin2A/b2 =
1, from which we can isolate r to get:

cos2A/a2 + sin2A/b2 =
1/r2

We can eliminate the squared trigonometric terms by recalling that:

(cos2A - sin2A) = (2cos2A - 1) = (1 - 2sin2A)
= cos2A and therefore:

cos2A = (1 + cos2A)/2

sin2A = (1 - cos2A)/2

Plugging these values back into the formula for 1/r2, we get:

(1/a2)(1 + cos2A)/2 + (1/b2)(1 - cos2A)/2 = 1/r2

We can now isolate the cos2A terms to get:

(1/b2 + 1/a2)/2 - (1/b2 - 1/a2)(cos2A)/2
= 1/r2

This looks surprisingly like our Mohr Circle derivation for stress, except we
derived that strictly from vector analysis without thinking of ellipses at all
(It's no accident). The only slightly puzzling feature is that we've reversed
the a and b terms and there is a minus sign in front of the trigonometric term.

The reason for the changes is clear at left. In this Mohr
Circle space, all radii are mapped as their inverse squares (called quadratic
extensions). Since a is bigger, its inverse square is smaller. All the
other rules of Mohr Circles still apply.

Now, what does h (red) represent? The radius of the Mohr Circle is (1/b2 - 1/a2)/2, so h
= ((1/b2 - 1/a2)/2)sin2A. What does that represent on the
ellipse?

We might suspect (rightly, it turns out) that h has something to do with
shear, as the corresponding coordinate does on the Mohr Circle for stress. Let's
begin with a circle and deform it into our ellipse as shown below:

On the left we have a circle with a tangent line P'Q'. On the circle the
radius is perpendicular to the tangent. On the right the diagram has been
flattened by a factor b/a. All the triangles on the left are similar, but not on
the right. Also the radius is no longer perpendicular to the tangent. The angle
C, the change in angle due to flattening, is a measure of shear. Dimensions in
the horizontal direction are unchanged. We have to do a tedious bit of trig
manipulation but it's perfectly straightforward.

Tan B' = OQ'/OP and Tan B = OQ/OP = (b/a)OQ'/OP = (b/a) Tan B'

Tan A = y/x = (b/a)y'/x = (b/a) Tan A'

Tan C = Cot(A+B) = (1 - Tan A Tan B)/(Tan A + Tan B)

Now:

Tan B = (b/a) Tan B',

Tan B' = Cot A' = 1/Tan A', and

Tan A' = (a/b)Tan A. Therefore:

Tan B = (b2/a2)/ Tan A. Thus,

Tan C =

(1 - Tan A Tan B)/(Tan A + Tan B) =

(1 - (b2/a2)Tan A/Tan A)/(Tan A + (b2/a2)/
Tan A) =

(1 - (b2/a2)/(Tan A + (b2/a2)/
Tan A) =

(a2 - b2)/(a2Tan A + b2/Tan A)
=

(a2 - b2)Tan A/(a2Tan2A + b2)
=

(a2 - b2)(Sin A/Cos A)/(a2Sin2A/Cos2A
+ b2) =

(a2 - b2)(Sin ACos A)/(a2Sin2A
+ b2Cos2A)

We recall that cos2A/a2 + sin2A/b2 =
1/r2. The bottom term in the formula is thus equal to a2b2/r2.
So we can now write:

Tan C = (a2 - b2)(Sin ACos A)/(a2b2/r2)

= r2(a2 - b2)/a2b2)(Sin
ACos A)

= r2(1/b2 - 1/a2)(Sin ACos A) Now Sin 2A
= 2 Sin A Cos A, so we have:

= (r2(1/b2 - 1/a2)/2)Sin 2A

= r2h

So: h = Tan C/r2 = (1/r2)Tan C.

Writing it this way allows us to see that the geometrical relationship
at left must hold.

Hyperbolas and Mohr Circles

The equation of a hyperbola is similar to that of an ellipse, differing only
in one sign:

x2/a2 - y2/b2 = 1

Note a couple of significant facts:

When x = 0, y2 = -b2. In other words, there
is no y-intercept.

When y = 0, x2 = a2 . The x-intercepts are
x = a and x = -a.

When x and y approach infinity, y/x approaches b/a. In other words, the
hyperbola approaches the lines y = xb/a and y = -xb/a. These lines are
called asymptotes.

We can proceed exactly as we did for the ellipse, and get the polar equation

r2cos2A/a2 - r2sin2A/b2 =
1, from which we can obtain

(1/a2 - 1/b2)/2 + (1/b2 + 1/a2)(cos2A)/2
= 1/r2

This formula leads to the Mohr Circle shown here.

Where the Mohr Circle crosses the vertical axis, 1/r2 is
zero and r is infinite. These points correspond to the asymptotes of the
hyperbola.

The Mohr Circle above is a little different from the ellipse only because we
chose the ellipse orientation so the long axis was always horizontal and
labelled a. For both the ellipse and hyperbola, the ratio b/a affects the
relative proportions in the x- and y- directions. It doesn't make sense to speak
of a "long" or "short" axis of a hyperbola, and the ratio
b/a can have any value. The larger b/a is, the more open the hyperbolas.

If the Mohr circle passes through the origin, the conic section is on the
borderline between an ellipse and a hyperbola, a parabola. However, it can't be
drawn. A Mohr Circle through the origin means 1/b2 = 0 or b infinite.
The equation for an ellipse reduces from x2/a2 + y2/b2 = 1
to x2/a2 = 1 or x = +/-a. The equation for a
hyperbola reduces from x2/a2 - y2/b2 = 1
to x2/a2 = 1 and again to x = +/-a. In the case of
the ellipse, we imagine we are in the middle of an ellipse with an infinitely
long b axis.

Stress, Strain, and Conic Sections

A circle deformed uniformly becomes an ellipse, so it's easy to see the
connection between strain and ellipses. But why do Mohr Circles work for stress?
We said nothing about ellipses in deriving the Mohr Circle for stress. Because there's a connection between stress and ellipses, too. In fact, all
tensor quantities (of which stress and strain are two examples) can be described
in terms of conic sections.

If we have two principal stresses, S1 and S2, we can draw an ellipse with
axes 1/S12 and 1/S22, and draw the ellipse x2/S12 +
y2/S22 = 1. For any given plane, we plot the pole to
the plane. The normal stress is the inverse square of the radius of the ellipse
(1/r2) along the pole direction. The shear stress is TanC/r2.
This ellipse is called the stress ellipse. For
situations where one stress is tensional and one compressional, the Mohr circle
corresponds to a hyperbola rather than an ellipse, so the proper name for stress
ellipses or hyperbolas is stress conic.